The root of the equation $|x^{2}-x-6|=x+2$ are given by:
- $-2,2,4$
- $0,1,4$
- $-2,1,4$
- $0,2,4$
Solve |x^2 - x - 6| = x + 2. Case 1: x^2 - x - 6 = x + 2 => x^2 - 2x - 8 = 0 => (x-4)(x+2) = 0 => x = 4, -2. Case 2: x^2 - x - 6 = -(x + 2) => x^2 - 4 = 0 => x = 2, -2. Combining gives -2, 2, 4.
To solve the modulus equation, we consider two cases where x + 2 is non-negative. For the first case, we write x squared minus x minus 6 equals x + 2, which simplifies to x squared minus 2x minus 8 equals 0; factoring gives (x minus 4)(x + 2) equals 0, yielding roots of 4 and -2. For the second case, we write x squared minus x minus 6 equals negative x minus 2, which simplifies to x squared minus 4 equals 0, yielding roots of 2 and negative 2. Both 4 and negative 2 satisfy the original equation, as does 2, making the complete set of valid roots negative 2, 2, and 4.